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Moving Charges and Magnetism question

2012 · Q92
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Moving Charges and Magnetism question

2012 · Q92

NEETPhysicsMoving Charges and MagnetismMCQ+4 / −1
A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α\alphaα-particle to describe a circle of same radius in the same field?
  1. A
    2 MeV
  2. B
    1 MeV
  3. C
    0.5 MeV
  4. D
    4 MeV
View written solutionFree

Correct answer: B

According to the principal of circular motion in a magnetic field

Fc=Fm⇒mv2R=qVB{F_c} = {F_m} \Rightarrow {{m{v^2}} \over R} = qVBFc​=Fm​⇒Rmv2​=qVB

⇒R=mvqB=PqB=2m.kqB \Rightarrow R = {{mv} \over {qB}} = {P \over {qB}} = {{\sqrt {2m.k} } \over {qB}}⇒R=qBmv​=qBP​=qB2m.k​​

Rα=2(4m)K′2qB{R_\alpha } = {{\sqrt {2\left( {4m} \right)K'} } \over {2qB}}Rα​=2qB2(4m)K′​​

RRα=KK′{R \over {{R_\alpha }}} = \sqrt {{K \over {K'}}} Rα​R​=K′K​​

but R=RαR = {R_\alpha }R=Rα​ (given)

∴\therefore∴ K = K' = 1 MeV

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